How do I develop a function that models the bacteria in relation to time? (data provided)

Question

Time (hours)	1	2	3	4	5
Bacteria weight (grams)	11.0	19.7	34.6	61.7	117.8

Answer 1

If you are asked to find a model without the use of a graphing calculator, you should look for the rate of change between the bacterial weight and the time. Since the weight does not change by the same amount each hour, your model would not be linear, but rather should be exponential.

An exponential model would have the form of y = a(b)^x where y is the weight of the bacteria after x hours, a is the initial weight of the bacteria and b is the growth rate.

First find the growth rate. The growth rate (b) can be found by dividing two consecutive y values (bacterial weight); this value should be relatively consistent for all data points if the model is accurate. For example, 19.7/11.0 = 1.79; 34.6/19.7 = 1.76; 61.7/34.6 = 1.78; 117.8/61.7 = 1.91

Your growth rate for the model equation could be one of those values as they are all relatively close or you could average them together (b = 1.81 in that case)

The final step would be to fine the initial weight of the bacteria (i.e. the weight at time = 0 hours). To do that, plug in a y and x value into the exponential equation to solve for a. For example, using a growth rate of 1.81, y =11 and x = 1, then 11 = a(1.81)¹ which would give you a = 6.08. You could repeat this process for all the points and average the a values to get a more precise number. The final model (using the previous calculation) would be y = 6.08(1.81)^x

To test the accuracy of the model, plug in the various times in for x and see how they correspond to the weights given. For this model, the weights calculated would be 11.0, 19.9, 36.1, 65.3, and 118.1 compared to 11.0, 19.7, 34.6, 61.7 and 117.8. This shows the model function is accurate.

If you are allowed to use a graphing calculator, it can find the exponential regression of data using the EXREG feature under the STAT→CALC of the Ti84 plus. Add the x and y values into L1 and L2, then calculate the regression equation. The exact function given is y = 6.03(1.80)^x with an r value of .9997 (very strong correlation). This is very close to the function determined by hand.

Sorry if this reply was too late. Hope it helps!

Answer 2

The %'s vary. Between the first 2 is up 79.09%

2nd 75.6

3rd 79.7

4th 90.9

I don't see a pattern.